-
Since 4/x is the base of a log function x is > 0.
The equation is then equivalent with
(4/x)2 = x2- 6
<=> 16 = x4- 6x2
let x2 = t
<=> 16 = t2- 6t
<=> t=-2 or t=8
<=> x = sqrt(8) (since x >0)
-
Calculate the first and second derivative of
y = x3. e-x
|
y' = 3x2. e-x- x3.e-x
y" = 6x.e-x - 3x2. e-x - 3x2. e-x + x3.e-x
y" = 6x.e-x - 6x2. e-x + x3.e-x
-
Calculate the derivative of
y = ln(tan(x/2))
|
1 1 1
y' = --------- .--------------------.---
tan(x/2) cos(x/2).cos(x/2) 2
1
= -----------------------
2 sin(x/2).cos(x/2)
1
= ------
sin(x)
-
Calculate the derivative of
y = ln(tan(x/2 + pi/4))
|
Analogous as in the previous exercise, we find
y' = 1/cos(x)
-
The function f(x) is given by
(ex-1)/x for all x not 0
1 for x = 0
Investigate if f(x) is continuous for x = 0
|
f(x) is continuous for x = 0
<=> lim f(x) = f(0)
0
<=> lim f(x) = 1
0
ex - 1
Now, lim f(x) = lim ------- =
0 0 x
ex
= lim ---- = 1
0 1
Hence f(x) is continuous for x = 0
-
Find
sin(x) + cos(x) - ex
lim ----------------------
0 ln(1+ x2)
|
With l'Hospitals rule
cos(x) -sin(x) - ex
= lim --------------------
0 2x/(1 + x2)
cos(x) -sin(x) - ex
= lim (1 + x2) . --------------------
0 2x
cos(x) -sin(x) - ex
= lim (1 + x2) . lim --------------------
0 0 2x
- sin(x) - cos(x) - ex
1. lim -------------------------- = -1
0 2
-
Given : f(x) = ln(e-2 + ex)
Prove that f(x) increases for all x.
What is the equation of the inverse function?
|
f'(x) = (e-2 + ex)-1 . ex > 0 for all x.
So, f(x) increases for all x.
The equation of the function f(x) is
y = ln(e-2 + ex)
<=>
ey = e-2 + ex
<=>
ex = e-2 - ey
<=>
x = ln(e-2 - ey)
The equation of the inverse function is
y = ln(e-2 - ex)
-
Solve
x(2ln(x)-1) + e(1/9) = (1 + e(1/9)) x(ln(x)-0.5)
|
Let y = x(ln(x)-0.5)
Then the given equation becomes
y2 + e(1/9) = (1 + e(1/9)) y
This is a quadratic equation in y. The roots are
y = 1 and y = e(1/9)
a)
y = 1
<=> x(ln(x)-0.5) = 1
<=> (ln(x) - 0.5) = 0
<=> ln(x) = 0.5
<=> x = sqrt(e)
b)
y = e(1/9)
<=> x(ln(x)-0.5) = e(1/9)
Taking ln of both sides, we have
<=> (ln(x) - 0.5).ln(x) = 1/9
<=> ln(x).ln(x) - 0.5 ln(x) - 1/9 = 0
This is a quadratic equation in ln(x). The roots are
ln(x) = 2/3 and ln(x) = -1/6
<=> x = e2/3 and x = e-1/6
-
Instead of calculating the limit directly, we calculate
ln lim x-x = lim ln(x-x)
0
-ln(x)
= lim ( -x.ln(x)) = lim ------
0 0 1/x
With l'Hospitals rule we have
1/x
= lim ------- = lim x = 0
1/(x2)
So,
-x -x
ln lim x = 0 => lim x = 1
0 0
-
Calculate the derivative of xx
|
d
We know that --(ln ( xx ) )
dx
1 d
= ----. --( xx)
xx dx
d
So, --( xx)
dx
x d
= x . --(ln ( xx )
dx
x d
= x .--(x.ln ( x ) )
dx
x
= x .(ln(x) + 1)
You can generalize this method to calculate the derivative of
g(x)
f(x)
-
Find
ln(ln(1+x4))
lim ---------------
0 ln(ln(1+x2))
|
With l'Hospitals rule
ln(1+x2) 1+x2 4x3
= lim ----------.------.------
0 ln(1+x4) 1+x4 2x
2 2 2
x ln(1+x ) 1+x
= 2 lim ----------.lim-----
0 4 0 4
ln(1+x ) 1+x
x2.ln(1+x2)
= 2 lim -------------
0 ln(1+x4)
x4.ln(1+x2)
= 2 lim --------------
0 x2.ln(1+x4 )
by properties of logarithm
ln(1+x2)(1/x2)
= 2 lim --------------------
0 ln(1+x4)(1/x4)
by properties of the number e
ln e
= 2 ------ = 2
ln e
-
Investigate the function
ex + 3 e-x
y = ln(-----------------)
ex + 1
|
The argument of ln is strictly positive. The domain of the function is R.
ex + 3 e-x
y = 0 <=> ------------- = 1
ex + 1
<=> ex + 3 e-x = ex + 1 <=> 3 = ex
<=> x = ln 3
ex - 6 - 3 e-x
y' = ... = ----------------------------
(ex + 3 e-x) (ex + 1)
e2 x - 6 ex - 3
= --------------------------
(e2 x + 3) (ex + 1)
let ex= u then y' = 0 <=> u2- 6 u - 3 = 0
___
<=> u = 3 + 2 V 3
__
<=> x = ln(3 + 2 V 3)
Investigation of the sign of y' tells that there is a minimum for
that x-value.
Since the domain is R, there are no vertical asymptotes
To find the horizontal asymptotes we calculate
ex + 3 e-x
lim ln(---------------)
+ infty ex + 1
x -x
e 3 e
= ln(lim -------- + lim -------- ) = ln (1 + 0 ) = 0
x x
e + 1 e + 1
y = 0 is a horizontal asymptote for x -> + infty
ex + 3 e-x
lim ln(---------------)
- infty ex + 1
x -x
e 3 e
= ln(lim -------- + lim -------- ) = ln (1 + infty) = + infty
x x
e + 1 e + 1
There is not a horizontal asymptote for x -> - infty
To find the oblique asymptotes y = ax + b we calculate
ex + 3 e-x
a = lim ln(---------------) / x
-infty ex + 1
With l'Hospitals rule we find
e2 x - 6 ex - 3
= lim -------------------------- = -1
-infty (e2 x + 3) (ex + 1)
b = lim f(x) + x
-infty
This limit is not so easy to calculate because of the presence of ln.
Therefore we calculate
b (f(x)+x)
e = lim e
-infty
ex + 3 e-x
= lim (--------------) ex
ex + 1
e2x+ 3
= lim (-----------) = 3
ex + 1
It follows that b = ln 3.
The oblique asymptote is y = -x + ln 3
-
Solve next system for all real solutions
ex + e-y2 = 1 (1)
e2x + sqrt( e- y2) = 1 (2)
|
Let X = ex and Y = sqrt( e- y2)
Then X > 0 and Y > 0 and 2.ln(Y) = - y2 < 0
Thus, 0 < Y < 1
With this substitution the system becomes
/
| X + Y2 = 1 (3)
|
| X2 + Y = 1 (4)
\
(3) - (4) gives
X - X2 + Y2 - Y = 0
<=> (X - Y) - (X2 - Y2) = 0
<=> (X - Y) ( 1 - X - Y ) = 0
<=> X = Y or Y = 1 - X
First take Y = 1 - X .
(4) becomes X2 + 1 - X = 1
This gives X = 0 or ( X = 1 and Y = 0 )
Both cases give no solution for the original system.
Now take X = Y
(3) becomes X2 + X - 1 = 0
The positive root is X = (sqrt(5) - 1)/2
So, X = Y = (sqrt(5) - 1)/2
x = ln ( (sqrt(5) - 1)/2 ) = -0.481
and e- y2 = Y2 = (3 - sqrt(5))/2
<=> - y2 = ln ( (3 - sqrt(5))/2 )
<=> y2 = ln ( 2/(3 - sqrt(5)) )
<=> y = sqrt ( ln ( 2/(3 - sqrt(5)) ) ) = 0.981
or y = - sqrt ( ln ( 2/(3 - sqrt(5)) ) ) = - 0.981